Work done by a constant force/ The elevator (ch6, p2,pg 188)

You are moving into an apartment and take the elevator to the 6th floor. Suppose your weight is 685 N and that of your belongings is 915N.

(b) How much work does the elevator do on you alone (without belongings) on the downward trip, which is also made at a constant velocity?

this is the answer

*To determine the normal force, we will again use the fact that the elevator is moving at a constant velocity and apply Newton's second law with the acceleration set to zero. Question, :uhh:why are we applying newton's law? to get normal force? and why are we getting normal force?

*then it continues..........Since the force exerted by the elevator and the displacement are in opposite directions on the downward part of the trip, the angle between them is Θ=180°, and so the work done by the force is negative. Question:uhh:, why is the angle 180? which angle? I don't know where the 180° come from

You want to find the normal force because that is (one of) the force that is doing work on you.
Gravity also does some work on you (positive work, you gain energy by moving down)
But the elevator also does work on you (negative work, it takes away the energy that you would have if you just fell that height)

The 180 degree angle is just another way of saying it's negative.
For work, you only want the component of the force in the direction of the displacement, so you take F*d*cos(θ) where θ is the angle between the Force and Displacement
(cos(180°)=-1)
In this case, the displacement vector (pointing downwards) and the normal force vector (pointing upwards) are in complete opposite directions (meaning there's 180 degrees between them). This means the work done is negative.